Find rate of change for velocity of water running out of a hole in a tank A vertical cylindrical tank of radius 10 inches has a hole of radius 1 inch in its base. The velocity $v$ with which the water contained runs out of the tank is given by $v^2=2gh$, where $h$ is the depth of the water and $g$ is the acceleration due to gravity. How rapidly is the velocity changing?
Using the given relation of $v^2=2gh$ and taking $g$ to be a constant,I got $\frac{dv}{dt}=\frac{g}{v}\frac{dh}{dt}$. 
What should be the next step? Hints would be appreciated. 
 A: You already have
$$\frac{dv }{dt} = \frac gv \frac{dh }{dt} \tag1$$
Next, let $R$ and $r$ be the radii of the cylinder and the hole, respectively.  The water volume in the cylinder is $V=\pi R^2 h$ and the rate of volume change is the rate of water flowing out, i.e. $-\pi r^2 v$, Then, establish
$$\frac{dV}{dt} = \pi R^2 \frac{dh}{dt} =-\pi r^2 v\tag2$$
Substitute (2) into (1) to obtain the rate of velocity change
$$\frac{dv }{dt} = -\frac{r^2}{R^2}g=-\frac g{100}$$
Note that the velocity decreases at a constant rate.
A: You should always consider all the data of your problem, in this case, you must make use of the radius (10 inches) of the cylinder and one inch at the base.
A: $v(h) = \sqrt{2gh}$
$\frac{d(v(h))}{dh} = \frac{d\sqrt{2gh}}{dh}$
You could take the constant out of the derivative since it's being multiplied.
$\frac{d(v(h))}{dh} = \sqrt{2}\frac{d({(gh)^{1/2}})}{dh}$
Hint: use the chain rule
Edit 1: The question probably asks you to find the rate of change of velocity with respect to height and not with respect to time. The phrasing makes it ambiguous.
